speed of balloon which is rising upwards with constant speed is
[tex]v_1 = 1.6 m/s[/tex]
speed of the camera which is thrown up from the ground
[tex]v_2 = 10.2 m/s[/tex]
the distance between camera and balloon when it is thrown from the ground
[tex]d = 2.60 m[/tex]
now let say after time "t" the camera is catch by her friend in the balloon
So the position of balloon and camera must be same
[tex]y_{balloon} = y_{camera}[/tex]
[tex]2.6 + 1.60 t = 10.2* t + \frac{1}{2}(-g)t^2[/tex]
[tex]2.6 = 8.6 *t - 4.9 t^2[/tex]
Solving above quadratic equation we got
[tex]t = 0.39 seconds[/tex]
So after t = 0.39 s her friend will get her camera on the balloon.
It takes 0.388 seconds for the camera to reach her
These are the formulas that we have to remember before solving the problem.
Speed is the rate of change of distance.
[tex]\large { \boxed {v = \frac{d}{t}}}[/tex]
v = speed ( m/s )
d = distance ( m )
t = time ( s )
Acceleration is the rate of change of velocity.
[tex]\large { \boxed {a = \frac{\Delta v}{t}}}[/tex]
a = acceleration ( m/s² )
Δv = change in speed ( m/s )
t = time ( s )
Let us now tackle the problem!
Given:
Speed of Air Balloon = u = 1.60 m/s
Initial Speed of Camera = vo = 10.2 m/s
Initial Distance of Passenger and Camera = d = 2.60 m
Gravitational Acceleration = g = 9.80 m/s²
Unknown:
Time Required = t = ?
Solution:
When camera reaches her :
displacement of camera = 2.60 + displacement of passenger
[tex]v_o ~ t - \frac{1}{2} ~ g ~ t^2 = 2.60 + u ~ t[/tex]
[tex]10.2 ~ t - \frac{1}{2} ~ 9.80 ~ t^2 = 2.60 + 1.60 ~ t[/tex]
[tex]4.9 ~ t^2 - 8.60 ~ t + 2.60 = 0[/tex]
To solve this quadratic equation, we can use the following formula:
[tex]t= \frac{-b - \sqrt{b^2-4ac}}{2a}[/tex]
[tex]t = \frac{8.60 - \sqrt{(-8.60)^2 - 4(4.9)(2.60)} }{2(4.9)}[/tex]
[tex]\boxed {t \approx 0.388 ~seconds}[/tex]
Grade: High School
Subject: Physics
Chapter: Kinematics
Keywords: Speed , Distance , Acceleration , Time , Velocity , Jet , Plane , TakeOff , Runway
